Ans -1
$\begin{bmatrix} 1 & 2 &3 & 4 & 5\\ 5& 1 & 2 & 3 & 4\\ 4& 5 & 1 &2 & 3\\ 3& 4 & 5 & 1 & 2\\ 2& 3 &4 & 5 & 1 \end{bmatrix}$
=$\begin{bmatrix} 15 & 15 &15 & 15 & 15\\ 5& 1 & 2 & 3 & 4\\ 4& 5 & 1 &2 & 3\\ 3& 4 & 5 & 1 & 2\\ 2& 3 &4 & 5 & 1 \end{bmatrix}$
[where $R_{1}\leftarrow R_{1}+R_{2}+R_{3}+R_{4}+R_{5}$]
=$15\begin{bmatrix} 1 & 1 &1 & 1 & 1\\ 5& 1 & 2 & 3 & 4\\ 4& 5 & 1 &2 & 3\\ 3& 4 & 5 & 1 & 2\\ 2& 3 &4 & 5 & 1 \end{bmatrix}$
=$15\begin{bmatrix} 1 & 1 &1 & 1 & 1\\ 14& 13 & 12 & 11 & 10\\ 4& 5 & 1 &2 & 3\\ 3& 4 & 5 & 1 & 2\\ 2& 3 &4 & 5 & 1 \end{bmatrix}$
[Here $R_{2}\leftarrow R_{2}+R_{3}+R_{4}+R_{5}$]
=$15\begin{bmatrix} 0 & 0 & 0 & 0 &1 \\ 1 & 1 &1 &1 & 10\\ -1&4 & -1 &-1 & 3\\ -1 & -1 & 4 &-1 &2 \\ -1 & -1 &-1 &4 & 1 \end{bmatrix}$
[Here $C_{1}\leftarrow C_{1}-C_{2}$
$C_{2}\leftarrow C_{2}-C_{3}$
$C_{3}\leftarrow C_{3}-C_{4}$
$C_{4}\leftarrow C_{4}-C_{5}$]
=$15\times 1\begin{bmatrix} 1 & 1 & 1 &1 \\ -1& 4 & -1 &-1 \\ -1 &-1 &4 &-1 \\ -1& -1 &-1 &4 \end{bmatrix}$
=$15\times 1\begin{bmatrix} 0& 0 & 0 &1 \\ -5& 5 &0 &-1 \\ 0 &-5 &5 &-1 \\ 0& 0 &-5 &4 \end{bmatrix}$
=$15\times 1\times 5\times 5\times 5\begin{bmatrix} -1& 1 &0 \\ 0 &-1 &1 \\ 0& 0 &-1 \end{bmatrix}$
=$15\times 1\times 5\times 5\times 5\begin{bmatrix} -1-\lambda & 1 &0 \\ 0 &-1-\lambda &1 \\ 0& 0 &-1-\lambda \end{bmatrix}$
=(-1-$\lambda$)(-1-$\lambda$)^2
$\lambda =-1$
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How eleminating column
$\begin{bmatrix} 0 &0 & 1\\ x& y &z \\ p& q &r \end{bmatrix}$
=$0\begin{bmatrix} y & z\\ q & r \end{bmatrix}$-$0\begin{bmatrix} x & z\\ p& r \end{bmatrix}$+$1\begin{bmatrix} x & y\\ p& q \end{bmatrix}$
=$1\begin{bmatrix} x & y\\ p& q \end{bmatrix}$
So, 1 row, and 1 column eleminated